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Current time:0:00Total duration:10:36

Video transcript

last few videos we looked at the different confirmations of cyclohexane and obviously cyclohexane is just one ring so it'd be a mono-cyclic compound the official IU PAC way of telling how many rings you have is how many cuts does it take to make an open chain alkane so if I if I just made a cut between those two carbons I have an open chain alkane and it took me one cut to get there so obviously cyclohexane is mono-cyclic what about if I had a cyclohexane ring and then I had another ring fused to it like that obviously there are two rings here so this would be a bicyclic compound once again if you were to use the IU pack way of figuring out how many rings you have you would make a cut here and a cut here and now you have an open chain alkane so it took you two cuts to get there so you would classify this as a bicyclic molecule now these are very easy simple compounds but for more complicated compounds you'll need to remember that I you PAC rule to figure out how many rings you actually have let's go ahead and redraw that that second molecule here and let's look at it in some more detail so I had a cyclohexane ring and then I had another ring fused to it like that now the carbons that are shared by both of these rings so this carbon and this carbon those are called your bridgehead carbons let's go ahead and write that the the bridge head carbons and when you're numbering your by cyclic compounds you want to start at the bridgehead carbon then number along the longest path then the second longest path and then finally the shortest path so let's go ahead and do that let's start with the top bridgehead carbon we'll make that number one and we're going to number along the longest path so the longest path would be going left so that's one two three four five and six and now the second longest path well we can just continue and there'll be seven and eight since that's the second longest path the third path the shortest path is usually the one between your bridgehead carbons there are no carbons between our bridgehead carbons so we're done in terms of numbering our compound the general the general formula for naming a bicyclic compound alright we've already determined this is a bicyclic compound so you would have you would have by cyclo here and then in brackets you would put the number of carbons in the longest path excluding the bridgehead carbon which I will represent here with X and then Y is the number of carbons of the second longest path excluding the bridgehead carbon and then Z is the number of carbons in the shortest path excluding the bridgehead carbon and then you finish everything off with the alkane so how many carbons are there in this molecule so for this example let's go ahead and name it we've already determined that it's a bicyclic compound so we'll go ahead and write by cyclo here and X again is the number of carbons in the longest path excluding the bridgehead carbon so we're going to find the longest path which is on the left here we're going to ignore the bridgehead carbon how many carbons were there 1 2 3 & 4 so we'll go ahead and put a 4 here alright why was the number of carbons in the second longest path well that was over here where we had this carbon this carbon again excluding the bridgehead carbon so that would get a 2 and finally the shortest path which was how many carbons are there between my bridgehead carbons and there of course zero carbons between my bridgehead carbons so if there are 0 carbons between your bridgehead carbons obviously your bridgit carbons are bonded directly together to which we can see in the dot structure finally how many carbons are there total there are 8 and we know that means octane so the final name for this molecule is by cyclo 4 to 0 octane let's do another one following our general formula here for naming a bicyclic compound so let's take a cyclohexane ring and let's make this a bridgehead carbon and we're actually going to put a carbon between our bridgehead carbons like that now this is kind of hard to see when it's drawn this way so usually you'll see it drawn a little bit differently usually you'll see it drawn with a little bit more three dimensionality to it so hopefully we can do that here so it would look something like this all right so let's let's find our bridgehead carbons here all right so we know that this is a bridgehead carbon this is a bridgehead carbon those correspond to these guys over here and how many cuts would it take to turn this into an open chain alkane well let's let's go ahead and look at the drawing on the left and if we cut it right here and we cut it here we can see it's now an open chain compound so it took two cuts so it's by cyclo so let's go ahead and and put those put those bonds back in there like that so it's a by cyclo compound so we can go ahead and start naming it as a bicyclic compound so this is by cyclo we need to go ahead and number it so let's start with our bridgehead carbon so this we'll start with this is our bridgit carbon find the longest path well it doesn't matter if I go left or right here since it's the same length each way around so I'll just go to the right so one two three four second longest path so I'm going to keep on going this way so five and six and then finally the shortest path which is the one carbon in between my two bridgehead carbons which would then get a seven so seven total carbons alright so when I'm naming this I need to put my brackets in here and I put the number of carbons in the longest path excluding the bridgehead carbon here so how many carbons were my longest path excluding the bridgehead here's one and here's two so I'm going to go ahead and put a two hidden here and then I go to my second longest path well that was over on this side and they're also two carbons on my second longest path obviously it's the same length as the other one so it's two two and then my shortest path how many carbons are there in my shortest path not counting my bridgehead carbon there's one so I go ahead and put a 1 here so I have by cyclo 2 2 1 0 and then my total number of carbons is 7 so this is a bycicle heptane molecules so by cyclo two-two-one heptane would be the official IU PAC name for this molecule this molecule turns up a lot in nature so much so that actually has its own common name this is also called nor boring so let's go ahead and write that here so this is a nor boring again a structure found often in nature alright let's do let's do one more let's do one more example here and let's make it look similar to the two that nor boring molecule so we'll have we'll have go like this alright so definitely definitely is a little bit tricky to draw these molecules here so let's put let's put some substituents on this molecule so i'm going to attempt to draw the same molecule over here and we're going to take a look at two different ways to number it and see which one turns out to be correct so here i have the exact same molecule on both sides here and let's see let's see how to do this so let's first find our bridgehead carbons so here are my bridgehead carbons right here now when I start numbering I could start at either one of those bridgehead carbons so if I identify my bridgehead carbons over on this one I could start numbering with with either of these two being number one so let's just make let's just make this number one here so that's number one I next go to my longest path so that would be going to the right here so this is giving me a two for this carbon a three for this carbon a four for this carbon a five for this car but next the second longest path well I can just keep going make that a six make this a seven and then finally my shortest path which is up here at the top so that would be eight total carbons for my parent name on the on the dot structure and left again the exact same dot structure this time I'm going to start with the opposite bridgehead carbons so I'm going to start with this bridgehead carbon and then follow the same rules so longest path excluding your so if all longest path which we to the right here so two three four five six seven and then eight so which one of these numbering systems is the correct one remember our goal is to get the lowest number possible for our substituents and if I look at the example on the right I have a methyl group in the seventh position and a methyl group in the 8th position the example on the left has a methyl group in the sixth position and a methyl group in the 8th position since I want to give my lowest number possible to my substituent I'm going to not number it the way on the right I'm going to go with the way on the left here so let's go ahead and start naming it we already know this is a bicyclic compound and we have two methyl groups one at six and one and eight so we can start naming it by saying six eight dimethyl and then we know this is a bicyclic compound so by cyclo and then in the brackets we're going to do the longest path first excluding the bridgehead carbon so the longest path is on the right how many carbons are there in the longest path there are three excluding the bridgehead carbons so we'll go ahead and put a three in here next the second longest path excluding the bridgehead carbon well that was these two right here so next there will be a two and then finally the shortest path excluding our bridgehead carbon there is only one carbon in our shortest path so we put that in there and then there were eight total carbon so it is octane so the final I u-pack name for this molecule is six eight dimethyl by cyclo three two one octane